JEE MAIN - Chemistry (2019 - 10th January Evening Slot - No. 20)
The ground state energy of hydrogen atom is – 13.6 eV. The energy of second excited state of He+ ion in eV is :
$$-$$ 6.04
$$-$$ 54.4
$$-$$ 27.2
$$-$$ 3.4
Explanation
(E)nth = (EGND)H . $${{{Z^2}} \over {{n^2}}}$$
E3rd (He+) = ($$-$$13.6eV) . $${{{2^2}} \over {{3^2}}}$$ = $$-$$ 6.04 eV
E3rd (He+) = ($$-$$13.6eV) . $${{{2^2}} \over {{3^2}}}$$ = $$-$$ 6.04 eV
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